When the numbers are given in equal intervals, another way is to calculate the differences of the given numbers. If the differences are not constant, it means that the function does not represent a straight line.
However, you can calculate the differences of the previously calculated differences. If the second difference is constant, the function is a quadratic. You can then continue the sequences to find f(9).
OpenStudy (anonymous):
How do i keep it going to find the answer for 9 hours though
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OpenStudy (xapproachesinfinity):
the second difference is constant 0.76
so it really is a quadratic
OpenStudy (anonymous):
so how do we use the quadratic to find the answer? im confused and need to show my work
OpenStudy (xapproachesinfinity):
there is a different method then the sequence
f(x)=ax^2+bx+c
f(0)=5.1
f(1)=3.03
f(2)=1.17
you got 3 equation in terms of a,b,c
solve for them
OpenStudy (xapproachesinfinity):
if you got the equation
you plug 9
for f(9)
OpenStudy (anonymous):
A=3.8 B=-2.45 and C=5.1
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OpenStudy (xapproachesinfinity):
no no
OpenStudy (anonymous):
I had to find that in the question previous to this one
OpenStudy (anonymous):
Are we doing it differently because we are plugging in the 9?
OpenStudy (xapproachesinfinity):
you have to find a,b , and c plug 0 for one equation
1 for the second
2 doe the third
system of three equation
OpenStudy (xapproachesinfinity):
f(0)=5.1=c
actually c is good!
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OpenStudy (xapproachesinfinity):
how did you find b and a
OpenStudy (xapproachesinfinity):
f(1)=3.03=a+b
f(2)=1.72=4a+2b
OpenStudy (xapproachesinfinity):
oh i forgot c
a+b+c=3.03
4a+2b+c=1.72
OpenStudy (xapproachesinfinity):
c=5.1
OpenStudy (anonymous):
Ok i've written what we have so far
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OpenStudy (xapproachesinfinity):
solve that for a and b
OpenStudy (anonymous):
4a+2b=-3.38?
OpenStudy (xapproachesinfinity):
yes!
now you have two equation
solve for a and b
OpenStudy (xapproachesinfinity):
i got a=0.38
b=-2.45
OpenStudy (anonymous):
thats what i have
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